Posts Tagged ‘line of action’

Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them.

From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F1 to the driven gear Force vector F2, then use that linking equation to develop a separate torque formula for each of the gears in the train.

We learned in the previous blog in this series that F1 and F2 travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value ϴ with respect to their Distance vectors D1 and D2. This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation:

F = [F1 × sin(ϴ)] – [F2 × sin(ϴ)]

where F is called a resultant Force vector, so named because it represents the force that results when the dead, or inert, weight that’s present in the resisting forceF2 cancels out some of the positive force of F1.

Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train.

We’ve been discussing torque and how it enables more power to be available to applications such as loosening tight nuts with a wrench. Now we’ll see how those same principles apply to another application, a simple gear train.

To review, the torque formula is,

Torque = Distance × Force × sin(ϴ)

where, Distance and Force are vector magnitudes and ϴ is the angle formed between them.

Referring to the gear train illustration above, we see that Force and Distance vectors are present, just as they had been in our previous wrench/nut example. But instead of torque being created by way of force that’s applied to a wrench, things are reversed, and it’s the torque that creates the force.

You see, in the wrench/nut example, the force applied to the wrench handle created torque on the nut. In our present gear train example, the torque applied to the motor shaft is created by an electric motor exerting pressure upon the motor shaft, which in turn exerts a force upon the driving gear teeth. The driving gear is also attached to this shaft, so torque causes the driving gear to rotate along with the motor. This rotation results in a force being exerted at the point where the teeth of the driving gear mesh with the teeth of the driven gear. In other words, in the wrench/nut example force created torque, while in the present example torque creates a force.

The gear train has a pivot point, as there was in our wrench/nut example, but this time it’s located at the center of the motor shaft rather than at the center of a nut. The pivot point in both examples is where the action takes place. The motor’s shaft and driving gear rotate around it, just as the wrench jaws and handle rotated around the nut’s pivot point.

In both examples, the Distance vectors extend out from the pivot points to meet up with the Force vector’s path. In the gear train example, this Force vector path is called a line of action, as introduced earlier in this blog series. This line of action passes through to the point where the driving and driven gear teeth mesh. The force acting upon that point causes the gears in the gear train to rotate, and as they turn mechanical energy is transferred from the motor to whatever machinery component is attached to the shaft of the driven gear. The powered component will then be able to perform useful work such as cutting lumber, mixing frosting for a cake, drilling holes in steel, or propelling vehicles.

You will note that there is an angle ϴ which exists between the Distance and Force vectors. Since we have a pivot point, a Force vector, a Distance vector, and an angle ϴ, we are able to apply the torque formula to gear trains exactly as we did in our wrench/nut example. We can then use that formula to calculate how torque is transmitted between gears in the train.

Next time we’ll examine the distance and force vectors in a simple gear train.

Last time we learned about forces generated when spur gear teeth mesh and move along a specific line of action. Today we’ll see them in movement.

Looking at the illustration below it might appear that there are three teeth in contact, but this isn’t the case. As the gears rotate, only two teeth make contact at any given time, although the third tooth comes very close. The actual point of contact between the teeth is represented by a black dot on the illustration below. This is where two opposing forces, F1 andF2, meet.

Now let’s animate the illustration to see how the line of action remains constant the entire time the gear teeth are in motion. By constant I mean that this imaginary line’s position and angle does not change relative to the gears throughout the course of their movement.

In the animation, the point of contact moves along the line of action as the gears turn. Each tooth’s involute profile ensures that smooth contact is maintained along the faces and flanks of the gear teeth. The involute profile’s unique shape facilitates opposing teeth remaining in constant contact along the line of action for the duration of their movement together.

If the gear tooth profile wasn’t involute in its shape, say for example it was square or triangular, the forces acting upon the meshed teeth during movement would vary in direction and intensity as a result of uneven contact between the teeth. For example, consider the square shaped tooth profile in the gear train below.

As the gears rotate, the pointed tip of one tooth strikes the flat face of another. As they continue to turn, the two flat faces of the teeth slap together, then the pointed tip of one tooth will strike the flat face of the other tooth, and so forth. The result is movement that is jerky and destructive. There would be excessive vibration and wear and tear on the teeth, resulting in rapid gear tooth erosion and decreased efficiency overall.

Next time we’ll introduce the gear ratio, a formula which allows us to alter the rotational speed of the driven gear in relation to that of the driving gear, something which comes in handy when designing things that require this differential.

Inertia. It’s the force that keeps us in bed after the alarm has rung. It seems to have a life of its own, and today we’ll see how it comes into play in keeping other stationary objects at rest.

Last time we identified a specific point of contact between spur gear teeth in a gear train and introduced the opposing forces, F1 and F 2, generated there. Today we’ll see what these forces represent, identifying one of them as inertia.

So where do these forces come from? They’re forces generated by different means that converge at the same point of contact, the point at which gear teeth mesh. They follow a very specific geometric path to meet there, an imaginary straight line referred to as the line of action.

F1 is always generated by a source of mechanical energy. In our locomotive example introduced earlier in this blog series that source is an electric traction motor, upon which a driving gear is mounted. When the motor is energized, a driving force F1 is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear.

Force F2 is not as straightforward to understand, because it’s not generated by a motor. Instead, it’s the resisting force that the weight of a stationary object poses against its being moved from an at-rest position, known as inertia. The heavier the object, the more inertia it presents with. Trains, of course, are extremely heavy, and to get them to move a great deal of inertia must be overcome. Inertia is also a factor in attempting to stop objects already in motion.

To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle. At their point of contact, the driving force of the motor, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2. If F1 is less than or equal to F2, then the train won’t leave the station.

Next week we’ll animate our static image and watch the interplay between gear teeth, taking note of the line of action during their movement.